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Stells [14]
3 years ago
15

Dolores made two isosceles right triangles by cutting the square along its diagonal. She says she can find the area of one of th

e triangles by using the formula for the area of the square, and then dividing the answer by 2. Do you agree? Explain.

Mathematics
1 answer:
kondaur [170]3 years ago
7 0

Answer:   Yes, I agree. The explanation is given below.


Step-by-step explanation:  As given in the question, a square ABCD is shown in the attached figure. Dolores cut the square along its diagonal BD and made two isosceles right angled triangles ΔABD and ΔCBD.

We need to show the area of triangles ABD and CBD are equal, and half of the area of square ABCD, i.e.,

area of ΔABD = are of ΔCBD = half of area of square ABCD.

Now, in ΔABD and ΔCBD, we have

AB = CD,

AD = BC

and BD is the common side.

Therefore, by sing SSS (side-side-side) postulate, we get

ΔABD ≅ ΔCBD.

So, area of ΔABD = area of ΔCBD.

Now, area of square ABCD =  area of ΔABD + area of ΔCBD

                                              = 2 × area of ΔABD

                                              = 2 × area of ΔCBD.

Thus, Doroles is absolutely correct in finding the area of any one of the triangles by using the formula for the area of the square and then dividing by 2.

Hence explained.



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The points (2, r) and (5, -9) lie on a line with slope – 2. Find the missing coordinate r.
Veseljchak [2.6K]

Answer:

-3

Step-by-step explanation:

5-2 is three

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-9-(-6) is -3

(2,-3)

3 0
3 years ago
Cos ( α ) = √ 6/ 6 and sin ( β ) = √ 2/4 . Find tan ( α − β )
Zina [86]

Answer:

\purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

Step-by-step explanation:

\cos( \alpha ) =  \frac{ \sqrt{6} }{6}  =  \frac{1}{ \sqrt{6} }  \\  \\  \therefore \:  \sin( \alpha )  =  \sqrt{1 -  { \cos}^{2} ( \alpha ) }  \\  \\  =  \sqrt{1 -  \bigg( {\frac{1}{ \sqrt{6} } \bigg )}^{2} }  \\  \\ =  \sqrt{1 -  {\frac{1}{ {6} }}}  \\  \\ =  \sqrt{ {\frac{6 - 1}{ {6} }}}   \\  \\  \red{\sin( \alpha ) =  \sqrt{ { \frac{5}{ {6} }}} } \\  \\  \tan( \alpha ) =  \frac{\sin( \alpha ) }{\cos( \alpha ) }  =  \sqrt{5}  \\  \\ \sin( \beta )  =  \frac{ \sqrt{2} }{4}  \\  \\  \implies \: \cos( \beta )  =   \sqrt{ \frac{7}{8} }  \\  \\ \tan( \beta )  =  \frac{\sin( \beta ) }{\cos( \beta ) } =  \frac{1}{ \sqrt{7} }   \\  \\  \tan( \alpha  -  \beta ) =  \frac{ \tan \alpha  -  \tan \beta }{1 +  \tan \alpha .  \tan \beta}  \\  \\  =  \frac{ \sqrt{5} -  \frac{1}{ \sqrt{7} }  }{1 +  \sqrt{5} . \frac{1}{ \sqrt{7} } }  \\  \\  =  \frac{ \sqrt{35} - 1 }{ \sqrt{7}  +  \sqrt{5} }  \\  \\  \purple{ \bold{ \tan( \alpha  -  \beta ) = 1.00701798}}

8 0
3 years ago
Which statement is true about the sum of (5y/y^2)+(y-4/5y) ? A: The denominator of the simplified sum is y2 + 5y. B: The denomin
lora16 [44]

Answer:

Dude the answer is B

Step-by-step explanation:

7 0
4 years ago
Find the equation of the line that goes through (-8,11) and is perpendicular to x= - 15. Write the equation in the form x = a, y
galben [10]

Answer:

y=11

Step-by-step explanation:

Hi there!

We want to find the equation of the line that passes through the point (-8, 11) and is perpendicular to x=-15

If a line is perpendicular to another line, it means that the slopes of those lines are negative and reciprocal; in other words, the product of the slopes is equal to -1

The line x=-15 has an undefined slope, which we can represent as 1/0, which is also undefined.

To find the slope of the line perpendicular to x=-15, we can use this equation (m is the slope):

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m_1 in this instance would be 1/0, so we can substitute it into the equation:

\frac{1}{0} *m_2=-1

Multiply both sides by 0

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So the slope of the new line is 0

We can substitute it into the equation y=mx+b, where m is the slope and b is the y intercept:

y=0x+b

Now we need to find b:

Since the equation passes through the point (-8,11), we can use its values to solve for b.

Substitute -8 as x and 11 as y:

11=0(-8)+b

Multiply

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So substitute into the equation:

y=0x+11

We can also write the equation as y=11

Hope this helps!

5 0
3 years ago
Slope = 4, passing through (-6, 8)
Mademuasel [1]

The point slope form is y - 8 = 4(x + 6) and the slope intercept form would be y = 4x + 32.

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y - 8 = 4(x + 6)

Now to find slope intercept form, solve for y.

y - 8 = 4(x + 6) ----> distribute the 4

y - 8 = 4x + 24 -----> add 8 to both sides

y = 4x + 32

7 0
3 years ago
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